# When 8.05 g of an unknown compound X was dissolved in 100. g of benzene (C6H6), the vapor pressure of the benzene decreased from 100.0 torr to 94.8 torr at 26 degrees C. What is the mole fraction and molar mass of X?

##### 1 Answer

I got

The reduction of vapor pressure is a **colligative property**. From Raoult's law, we have:

#P_j = chi_jP_j^"*"# ,where

#P_j# is the partial pressure above the solution from component#j# ,#P_j^"*"# is the pure vapor pressure of#j# by itself, and#chi_j# is the mol fraction of#j# in the solution.(We consider

#bbj# to be thesolventand#bbi# to be thesolute.)

But mol fractions are always

#P_j = (1 - chi_i) P_j^"*"#

Since

This reduction is calculated as follows. Define

#color(green)(DeltaP_j) = chi_jP_j^"*" - P_j^"*"#

#= (chi_j - 1)P_j^"*"#

#= (1 - chi_i - 1)P_j^"*"#

#= color(green)(-chi_iP_j^"*")#

Since *reduction in vapor pressure* is of course *negatively*-signed. From this equation then, we can determine the mol fraction of what was dissolved in benzene.

Plug

#(94.8 - 100.00) "torr" = -chi_i ("100.00 torr")#

#=> color(blue)(chi_i = 0.052)#

Now that we have the mol fraction of *molar* mass of *mols* of

#0.052 = (n_X)/(n_X + n_"Ben")#

#=> 0.052n_X + 0.052n_"Ben" = n_X#

#=> (0.052 - 1)n_X = -0.052n_"Ben"#

#=>n_X = 0.0549n_"Ben"#

But we know the molar mass of benzene; it's

#"100.0 g Benzene" xx "1 mol"/"78.1134 g" = "1.2802 mols Benzene solvent"#

Therefore, the mols of

#n_X = 0.0549("1.2802 mols") = "0.0703 mols X"#

This gives a molar mass for

#color(blue)(M_(m,X)) = "8.05 g X"/"0.0703 mols X" = color(blue)("114.51 g/mol")#